Second Derivative Test,

Graphical Connections

Section 4.3b

Do Now: #30 on p.204 (solve graphically)

21 2 4y x x x

1 2 4

(a) Local Maximum at

2x

(b) Local Minimum at

4x

(c) Points of Inflection:

1,x 1.634,3.366x

y

Second Derivative Test for Local Extrema

Let’s see these with the graphs of the squaringfunction and the negative squaring function…

1. If and , then has a local maximum at x = c.

0f c 0f c f

2. If and , then has a local minimum at x = c.

0f c 0f c f

Another Quick ExampleFind the extreme values of 3 12 5f x x x

2 23 12 3 4f x x x

2 12 0f

6f x x

2 12 0f

Critical Points???

x = 2, –2

Support our answers graphically???

has a local max. at x = –2

has a local min. at x = 2f

f

Let 3 24 12f x x x 1. Identify where the extrema of occur.

2. Find the intervals on which is increasing and the intervals on which is decreasing.

3. Find where the graph of is concave up and where it is concave down.

4. Sketch a possible graph for .

f

ff

f

f

Let 3 24 12f x x x is continuous since exists.

The domain of is all reals, so the domain of is the same.

Thus, the critical points of occur only at the zeros of .

24 3f x x x CP at x = 0, 3

Intervals

Sign of

Behavior of

x < 0

–Dec.

0 < x < 3

–Dec.

x > 3

+Inc.

Use first derivative test: Local Min. at x = 3

Decreasing on , Increasing on

f f f f

f f

f f

,0 , 0,3 3,

Let 3 24 12f x x x

The zeros of (IP) are at x = 0, 2

Intervals

Sign of

Behavior of

x < 0

+Conc. up

0 < x < 2

–Conc. down

x > 2

+Conc. up

Concave up on ,

212 24f x x x

Concave down on

12 2x x f

f

f

,0 , 2,

0,2

Let 3 24 12f x x x Summarizing info. from both tables:

x < 0

Decreasing

Concave up

0 < x < 2

Decreasing

Concave down

2 < x < 3

Decreasing

Concave up

x > 3

Increasing

Concave up

One possibility for the graph of :

x = 0

x = 2

x = 3

f

Let 3 24 12f x x x

One possibility for the graph of :

x = 0

x = 2

x = 3

Note: We are able to recover almost everything about adifferentiable function by examining its first derivative…

We cannot determine how to place the graph in the x-y plane(vertically) to position the graph we would need only thevalue of at one point!!!f

f

A function is continuous on , , , and derivatives have the following properties:

x –2 < x < 0

+

+

x = 0

does not exist

does not exist

0 < x < 2

–

+

x = 2

0

0

2 < x < 4

–

–

1. Find where all absolute extrema of occur.

2. Find where the points of inflection of occur.

3. Sketch a possible graph of .

f : 2, 4D 2 5f 4 1f

f f

f

f

f

A function is continuous on , , , and derivatives have the following properties:

x –2 < x < 0

+

+

x = 0

does not exist

does not exist

0 < x < 2

–

+

x = 2

0

0

2 < x < 4

–

–

f : 2, 4D 2 5f 4 1f

f f

Absolute Maximum occurs at x = 0(cannot determine its value)

Point of Inflection at x = 2

Absolute Minimum of 1 at x = 4